# Oxygen is contained in a pisto

Oxygen is contained in a piston-cylinder system initially at atemperature of 40 ^{o}C and a pressure of 150 kPa. Thepiston is 1 m in diameter and initially 20 cm above the base of thecylinder. Heat is added at constant pressure until the piston hasmoved 30 cm, so it’s now at 50 cm.

I found the mass of the system to be 0.2898 kg, and T2 =783K.

a) Using the table of properties of oxygen as an ideal gas,calculate the heat added to the system per unit mass, in kJ/kg

b) Repeat this calculation assuming that the specific heat as afunction of temperature is linear over this range of temperatures(ie, assuming the change in temperature is “small”)

c) Repeat this calculation without this assumption (ie, directintegration of empirical relation)

Answer:

Assume a piston-cylinder arrangement,with a piston diameter 1m

area of the piston ( A)0.785* 1^{2} =.785 m^{2}

at an initial condition temperatureT1 273 + 40 313 K

Pressure P_{1}= 150 kPa ,Volume V_{1}= A * L_{1} = .785 * 0.2= 0.157m^{3} ( L_{1} = .2 m GIVEN)

Here in this system heat is added atconstant pressure so initial pressure P_{1} = finalpressure P_{2} = 150 kPa

as OXYGEN is assume be an ideal gasso , C_{p} = 1.005 KJ/Kg and C_{v} = .718 KJ/Kg (Oxygen is a diatomic gas so we assume this value )

mass of the system (m)= 0.2898 kg

final temp. T_{2} = 783 K

final volume V_{2} = A*L_{2} = .785 * 0.5 = 0.3925 m^{3}

** a.**Heat added to the system per unit mass ( Q) =C

_{p}*T=1.005 * (783 – 313) = 472.35 KJ/Kg ( answer for a)

** b.**as the change of the temperature is small so we assume thatT = 0

As per 1st law of thermodynamics we know dQ = dU +dW

as T is equal to zero so dU = 0

so , dQ = dW = pdv = p (V_{2 }– V_{1} ) = 150 *( 0.3925-0.157) = 35.325 kJ / Kg

heat added for the mass m is = 35.325 * 0.2898 = 10.237 KJ (answer for b)

** c.**assuming that the specific heat as a function of temperature islinear over this range of temperatures so C

_{p}= a + bT (a& b are constant )

for initial condition for the table of O_{2}C_{p} 0.92 KJ/Kg

so , 0.92 = a + 313b ( equation 1)

for final condition for the O_{2} tableC_{p}1.046 KJ/Kg

so , 1.046= a + 783b ( equation 2)

now from this 2 equation we get the value of **a =0.833 **and **b= 2.766*10 ^{-4}**

so the equation become **C _{p =} 0.833 + ( 2.766*10^{-4})T**

**SO** heat added to this system by this process isper unit mass (dQ) = C_{p}T

total heat added ( Q ) = C_{p} T = (0.833 + (2.766*10^{-4})T) T

=**(**0.833( T _{2 }– T_{1}) + ( ( 2.766*10^{-4})/ 2 ) * (T_{2}^{2} -T_{1}^{2})**)**

**put** **T _{1 =} 313 andT_{2} = 783 then we get**

**Q** = 0.833 ( 783 – 313 ) + ( 1.383 *10^{-4}) ( 783^{2} – 313^{2} )

= 391.51 + 71.24 = 462.75 KJ/Kg

Total heat added for the system mass (m) = 462.75 * 0.2898 =134.1 KJ ( answer for C)